3.215 \(\int \frac {\cos ^m(c+d x) (B \cos (c+d x)+C \cos ^2(c+d x))}{(b \cos (c+d x))^{4/3}} \, dx\)

Optimal. Leaf size=173 \[ -\frac {3 B \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);\cos ^2(c+d x)\right )}{b d (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+5);\frac {1}{6} (3 m+11);\cos ^2(c+d x)\right )}{b d (3 m+5) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

[Out]

-3*B*cos(d*x+c)^(1+m)*hypergeom([1/2, 1/3+1/2*m],[4/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/b/d/(2+3*m)/(b*cos(d*x+c
))^(1/3)/(sin(d*x+c)^2)^(1/2)-3*C*cos(d*x+c)^(2+m)*hypergeom([1/2, 5/6+1/2*m],[11/6+1/2*m],cos(d*x+c)^2)*sin(d
*x+c)/b/d/(5+3*m)/(b*cos(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {20, 3010, 2748, 2643} \[ -\frac {3 B \sin (c+d x) \cos ^{m+1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);\cos ^2(c+d x)\right )}{b d (3 m+2) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}-\frac {3 C \sin (c+d x) \cos ^{m+2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+5);\frac {1}{6} (3 m+11);\cos ^2(c+d x)\right )}{b d (3 m+5) \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^m*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(4/3),x]

[Out]

(-3*B*Cos[c + d*x]^(1 + m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d
*(2 + 3*m)*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2]) - (3*C*Cos[c + d*x]^(2 + m)*Hypergeometric2F1[1/2, (5
+ 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*(5 + 3*m)*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^
2])

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rubi steps

\begin {align*} \int \frac {\cos ^m(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(b \cos (c+d x))^{4/3}} \, dx &=\frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac {4}{3}+m}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=\frac {\sqrt [3]{\cos (c+d x)} \int \cos ^{-\frac {1}{3}+m}(c+d x) (B+C \cos (c+d x)) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=\frac {\left (B \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{-\frac {1}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}+\frac {\left (C \sqrt [3]{\cos (c+d x)}\right ) \int \cos ^{\frac {2}{3}+m}(c+d x) \, dx}{b \sqrt [3]{b \cos (c+d x)}}\\ &=-\frac {3 B \cos ^{1+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (2+3 m);\frac {1}{6} (8+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (2+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {3 C \cos ^{2+m}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (5+3 m);\frac {1}{6} (11+3 m);\cos ^2(c+d x)\right ) \sin (c+d x)}{b d (5+3 m) \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 140, normalized size = 0.81 \[ -\frac {3 \sqrt {\sin ^2(c+d x)} \csc (c+d x) \cos ^{m+2}(c+d x) \left (B (3 m+5) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+2);\frac {1}{6} (3 m+8);\cos ^2(c+d x)\right )+C (3 m+2) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{6} (3 m+5);\frac {1}{6} (3 m+11);\cos ^2(c+d x)\right )\right )}{d (3 m+2) (3 m+5) (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^m*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/(b*Cos[c + d*x])^(4/3),x]

[Out]

(-3*Cos[c + d*x]^(2 + m)*Csc[c + d*x]*(B*(5 + 3*m)*Hypergeometric2F1[1/2, (2 + 3*m)/6, (8 + 3*m)/6, Cos[c + d*
x]^2] + C*(2 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, (5 + 3*m)/6, (11 + 3*m)/6, Cos[c + d*x]^2])*Sqrt[Sin[c
+ d*x]^2])/(d*(2 + 3*m)*(5 + 3*m)*(b*Cos[c + d*x])^(4/3))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) + B\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{m}}{b^{2} \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)*(b*cos(d*x + c))^(2/3)*cos(d*x + c)^m/(b^2*cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^m/(b*cos(d*x + c))^(4/3), x)

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maple [F]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \frac {\left (\cos ^{m}\left (d x +c \right )\right ) \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)

[Out]

int(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{\left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(4/3),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*cos(d*x + c)^m/(b*cos(d*x + c))^(4/3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^m\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^m*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(4/3),x)

[Out]

int((cos(c + d*x)^m*(B*cos(c + d*x) + C*cos(c + d*x)^2))/(b*cos(c + d*x))^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \cos ^{m}{\left (c + d x \right )}}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**m*(B*cos(d*x+c)+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(4/3),x)

[Out]

Integral((B + C*cos(c + d*x))*cos(c + d*x)*cos(c + d*x)**m/(b*cos(c + d*x))**(4/3), x)

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